Question

A battery has an emf of 1.50V. If you short circuit the battery with a wire with resistance 8.0Ω, a voltage of 1.12V is measured across the battery. What is the internal resistance of the battery?

Answer 1

Explanations & Calculations

• By Ohm's law to the wire $\to$ like an external resistance to the battery,

$\qquad\qquad \begin{aligned} \small i&=\small \frac{V}{R}=\frac{1.12\,V}{8\,\Omega}=0.14\,A \end{aligned}$

• By Kirchhoff's law to the entire circuit the current can be found again and you what's next,

$\qquad\qquad \begin{aligned}\\ \small E-ir-iR&=\small 0\\ \small i&=\small \frac{E}{(R+r)}\\ &=\small \frac{1.50\,V}{(8\,\Omega+r)} \end{aligned}$

• You know the current that flows in the circuit already and that value can be substituted in the last equation obtained to get the final answer.